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Bài 1:
\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)
Bài 2:
\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)
\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)
Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)
Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
\(\lim\limits_{x\to-\infty}\frac{x^4+2x^3+x-1}{x^2+3x-1}=\lim\limits_{x\to-\infty}\frac{x^2\left(x^2+3x-1\right)-x\left(x^2+3x-1\right)+4x^2-1}{x^2+3x-1}\)
\(\lim\limits_{x\to-\infty}\left(x^2-x+\frac{4x^2-1}{x^2+3x-1}\right)=\lim\limits_{x\to-\infty}\left(x^2-x+\frac{4-\frac{1}{x^2}}{1+\frac{3}{x}-\frac{1}{x^2}}\right)=\lim\limits_{x\to-\infty}\left(x^2-x\right)+4=+\infty\)
\(=\frac{\left|x\right|\sqrt{1+\frac{2}{x}}+3x}{\left|x\right|\sqrt{4+\frac{1}{x^2}}-x+3}=\frac{-x\left(\sqrt{1+\frac{2}{x}}-3\right)}{-x\left(\sqrt{4+\frac{1}{x^2}}+1+\frac{3}{x}\right)}=\frac{1-3}{2+1+0}=...\)
Lời giải:
\(\lim\limits_{x\to-\infty}\frac{3x^2-5x+2}{2x-3}=\lim\limits_{x\rightarrow-\infty}\frac{\frac{3}{2}x\left(2x-3\right)-\frac{1}{4}\left(2x-3\right)+\frac{5}{4}}{2x-3}=\lim\limits_{x\rightarrow-\infty}\left(\frac{3}{2}x-\frac{1}{4}+\frac{5}{4\left(2x-3\right)}\right)=\lim\limits_{x\to-\infty}\left(\frac{3}{2}x-\frac{1}{4}\right)=-\infty\)
Lời giải:
\(\lim\limits_{x\to-\infty}\frac{x^2-3x-2}{x^3+2x+1}=\lim\limits_{x\to-\infty}\frac{\frac{1}{x}-\frac{3}{x^2}-\frac{2}{x^3}}{1+\frac{2}{x^2}+\frac{1}{x^3}}=\frac{0}{1}=0\)